USA IMO, 2003

Extend segment $DM$ through $M$ to $G$ such that $FG \parallel CD$.

Then $MF = MC$ if and only if quadrilateral $CDFG$ is a parallelogram, or, $FD \parallel CG$. Hence $MC = MF$ if and only if $\angle GCD = \angle FDA$, that is, $\angle FDA + \angle CGF = 180^\circ$.

Because quadrilateral $ABED$ is cyclic, $\angle FDA = \angle ABE$. It follows that $MC = MF$ if and only if
$$180^\circ = \angle FDA + \angle CGF = \angle ABE + \angle CGF,$$
that is, quadrilateral $CBFG$ is cyclic, which is equivalent to
$$\angle CBM = \angle CBG = \angle CFG = \angle DCF = \angle DCM.$$
Because $\angle DMC = \angle CMB$, $\angle CBM = \angle DCM$ if and only if triangles $BCM$ and $CDM$ are similar, that is
$$\frac{CM}{BM} = \frac{DM}{CM},$$
or $MB \cdot MD = MC^2$.

We first assume that $MB \cdot MD = MC^2$. Because $\frac{MC}{MD} = \frac{MB}{MC}$ and $\angle CMD = \angle BMC$, triangles $CMD$ and $BMC$ are similar. Consequently, $\angle MCD = \angle MBC$. Because quadrilateral $ABED$ is cyclic, $\angle DAE = \angle DBE$. Hence
$$\angle FCA = \angle MCD = \angle MBC = \angle DBE = \angle DAE = \angle CAE,$$
implying that $AE \parallel CF$, and so $\angle AEF = \angle CFE$. Because quadrilateral $ABED$ is cyclic, $\angle ABD = \angle AED$. Hence
$$\angle FBM = \angle ABD = \angle AED = \angle AEF = \angle CFE = \angle MFD.$$
Because $\angle FBM = \angle DFM$ and $\angle FMB = \angle DMF$, triangles $BFM$ and $FDM$ are similar. Consequently, $\frac{FM}{DM} = \frac{BM}{FM}$, or $FM^2 = BM \cdot DM = CM^2$. Therefore $MC^2 = MB \cdot MD$ implies $MC = MF$.

Now we assume that $MC = MF$. Applying Ceva's Theorem to triangle $BCF$ and cevians $BM, CA, FE$ gives
$$\frac{BA}{AF} \cdot \frac{FM}{MC} \cdot \frac{CE}{EB} = 1,$$
implying that $\frac{BA}{AF} = \frac{BE}{EC}$, so $AE \parallel CF$. Thus, $\angle DCM = \angle DAE$. Because quadrilateral $ABED$ is cyclic, $\angle DAE = \angle DBE$. Hence
$$\angle DCM = \angle DAE = \angle DBE = \angle CBM.$$
Because $\angle CBM = \angle DCM$ and $\angle CMB = \angle DMC$, triangles $BCM$ and $CDM$ are similar. Consequently, $\frac{CM}{DM} = \frac{BM}{CM}$, or $CM^2 = BM \cdot DM$.